March 22, 2020
From Ebbinghaus, Flum, and Thomas’ Mathematical Logic, chapter 2, exercise 1.3:
Let be given. For such that show that there is a point in the closed interval such that . Conclude from this that , and hence also, are uncountable. (Hint: By induction define a sequence of closed intervals such that and use the fact that .)
Define . For any , if , let and define
For any , we can see that for any value of , we have , so Therefore, if , then for any .
As there is some , and for this , we have (and ) and for any , so for any given , then is not surjective onto (and ). Therefore, (and ) is not countable.